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27. The second polar of a point on the Steinerian touches the Hessian at the corresponding point.

A convenient form of the equation of the Hessian is [50], from which it follows that the degree of the Hessian is 4(n-2); accordingly since A is a point on the Hessian of (1), the highest power of a is the (4n-9)th. Now if we write down the expressions of a, b, ..., retaining only the highest power of a, as is done is § [51], it will be found that the coefficient of this term is

A d'u ̧] d8'= 2A (38v,+v,),

where is a constant. Hence the tangent plane to the Hessian at A is 38v,+v, = 0.

28. Let (, n,, w) be the coordinates of a point P on the Hessian, (, u, v, w) those of the corresponding point Q on the Steinerian, then the conditions that the polar quadric of P should have a conic node at Q are

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with three similar equations. The elimination of (X, μ, v, gives the equation of the Hessian, whilst that of (E, n, s, w) gives the equation of the Steinerian. Hence the degree of the latter is 4 (n − 2)3.

29. The polar plane of a point on the Hessian touches the Steinerian at the corresponding point.

Let us take to be the tangent plane to the Hessian at A; then v, 0. Let p be the coordinate of A, s that of D; let (p+, n,, 0) be those of a point P on the plane ABC near A; and let (λ, μ, v, s +) be those of the point Q which corresponds to P, where §, n, S, λ, μ, v, o are small quantities whose squares and products are to be neglected. Hence the first of (3) becomes

aλ+ h ̧μ+gv + 1 (s+w) = 0............................... (4), where the suffix denotes the values of the quantities at A. Now

a=n (n-1) p", h=(n-1) Ip",

!=(n−1)Kp"-, 1=0,

accordingly (4) becomes

nλ + Iμ + Kv = 0..............

(5),

which shows that the point Q lies on the plane

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na + 18+ Ky = 0............. ....(6), which is the polar plane of A. And since P is any point on the Hessian near A, (6) is the tangent plane to the Steinerian at D.

30. The first polars of the primitive curve, with respect to any point in the tangent plane to the Steinerian at D, touch the line AD at A, which joins D with the corresponding point A on the Hessian.

The first polar of (1) with respect to (§, n, 5, w) is

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which shows that the point (§, n, S, w) must lie on the tangent plane to the Steinerian at D.

The tangent plane to (7) at A is

{(n − 1) I§ + 2Ln+ M2 } ẞ + {(n − 1) K§ + Mŋ+2NY } y=0, which shows that the tangent plane at A to (7) passes through the line AD.

31. The class of the Steinerian is 4 (n−1)3 (n − 2).

Let O and O' be two points on a fixed line L in the tangent plane at D to the Steinerian. Then the first polars of the primitive curve with respect to O and O' intersect in a curve of degree (n−1)', which passes through the point A on the Hessian which corresponds to D. Hence the points in which this curve cuts the Hessian correspond to the points of contact of the tangent planes to the Steinerian which pass through the line L; and since their number is 4 (n−1)3 (n−2), this is the class of the Steinerian.

32. If the polar quadric of A consists of a pair of planes, then A is a conic node on the Hessian.

Choose the tetrahedron of reference so that CD is the line of intersection of the two planes. Then in (1) K=M=N=0;

41-9

whence it can be easily shown that the coefficient of a in the Hessian vanishes, and therefore A is a conic node.

33. We have shown in [60] that the Hessian has 10 (n-2) conic nodes. Now every point on the line CD corresponds to the conic node at A on the Hessian; hence CD is a line lying in the Steinerian. Accordingly the Steinerian possesses 10 (n-2) lines lying in the surface. By forming the equation of the nodal cone at A to the Hessian, it can be shown that the second polar of C (which may be any arbitrary point on CD) touches the nodal cone; and also that the polar plane of A touches the Steinerian along the line CD. Hence this line is not a torsal line, but a singular line analogous to a trope, since the tangent plane is fixed.

34. The class of the Hessian is 4 (n − 2) (11n2- 52n+61). The class m is determined from the equation

m=v (v-1)'-2 C,

and v = 4 (n − 2), C=10 (n − 2)3, which gives the result.

Fledborough Hall,

Holyport, Berks.

25 Feb. 1916.

ON THE PROOF OF ALGEBRAIC THEOREMS BY GEOMETRY.

By A. B. BASSET, M.A., F.R.S.

1. THIS paper is supplementary to the preceding one; and its object is to give geometrical proofs of certain theorems in the algebra of the binary and ternary quantic. References to the previous paper will be enclosed in square brackets.

There is an ancient precedent for this method. Euclid and his cotemporaries knew very little about Arithmetic and Algebra, and they were therefore obliged to prove the formulæ they required by Geometry. Their successors extended these methods, and Diocles invented a geometrical construction for finding the cube root of a number; but a serious impediment to the practical use of these methods arose from the fact that the art of constructing mechanical instruments for tracing curves was then in its infancy. At the present day a cissoid, a conchoid or a limaçon can be drawn with the same accuracy as a circle.

2. It is often convenient to write a quantic in the binary form

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If the coefficients are constants, u is a polynomial; if they are binary quantics of (3, y) of proper degrees, u is the trilinear equation of a plane curve of degree n; whilst if they are ternary quantics of (3, y, d), u is the quadriplanar equation of a surface.

When u is the trilinear equation of a plane curve, its discriminant with respect to a is of the form RT'S', where R is the equation of the tangents drawn from the vertex A of the triangle of reference to the curve, T is the equation of the lines joining A to each node, and S is that of the lines joining A to each cusp.

When u is the quadriplanar equation of a surface, its discriminant with respect to a is the equation of the tangent cone from the vertex A of the tetrahedron of reference to the surface; and it does not in general degrade into factors. If, however, the surface possesses a nodal and a cuspidal curve, the discriminant is of the above form, where R is the proper tangent cone, whilst T and S are respectively the cones, vertex A, which stand on the nodal and cuspidal

curves.

The simplest geometrical interpretation of a binary quantic is that it represents a number of straight lines, lying in a plane, which pass through a point; but there are two other interpretations. If a is a variable parameter and the coefficients of a are straight lines, the discriminant of u with respect to a is a unicursal plane curve. If the coefficients are planes, the discriminant is a developable surface whose reciprocal polar is a unicursal twisted curve. Hence the algebra of the binary quantic has important applications in the theory of unicursal plane and twisted curves and developable surfaces.

A ternary quantic of (a, ß, y) may be expressed in the form

v = (a', (3', 1)".......................................................... (2)

by writing a=a/r, B'=B/r; and if the coefficients of (a', B') are binary quantics of (y', ') of proper degrees, (2) represents the quadriplanar equation of a surface. Hence the geometry of surfaces may sometimes be employed with advantage to investigate the properties of ternary quantics.

3. The discriminant of the product of two binary quantics is equal to the product of their discriminants multiplied by their square of the eliminant.

Let u, u' be the trilinear equations of two curves of degree n, n'. Then if u, u' be written in the binary form (1) the discriminant A of the product uu' must be obviously equal to the tangents drawn from A to the compound curve, multiplied by the square of the lines E joining A to each node. But if u, u are anautotomic curves, the points of intersection of u and u' are the only nodes the compound curve possesses; hence

A = DD'E'

where D, D' are the tangents drawn from A to u, u' respectively, and are therefore equal to their discriminants with respect to a; whilst the lines E are obtained by eliminating a between u and u'. Hence E is the eliminant.

In like manner the discriminant of the product of any number of binary quantics u,, Uz, Uz, ..., is

D1, D2, ..., E',,, E',
E'...
239 ***

where D, is the discriminant of u,, and E,, the eliminant of u, and u,.

then

4. If a binary quantic be written in the form

U1 = (μ12 u1, ..., u„Čα, 1)”

n

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where AA are the discriminants of U, U-1

n-1

·(4),

The first term is accounted for by collecting together all the terms of the discriminant which contain u; hence the second term must be equal to the discriminant of the product αυ But by §3, this is equal to the discriminant A-, of U multiplied by the square of the lines joining 4 to the points where a cuts U.,, that is to u

2-1'

n-1

n-1'

In like manner it can be shown that

n-1

.(5),

n-1

where A is the equation of the tangents drawn to the curve from a point on itself.

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